In this post, we are going to solve the Search a 2D Matrix Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.
Problem
Write an efficient algorithm that searches for a value target in an m x n integer matrix matrix. This matrix has the following properties:
- Integers in each row are sorted from left to right.
- The first integer of each row is greater than the last integer of the previous row.
Example 1:
Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3 Output: true
Example 2:
Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13 Output: false
Constraints:
m == matrix.lengthn == matrix[i].length1 <= m, n <= 100-104 <= matrix[i][j], target <= 104
Now, let’s see the leetcode solution of Search a 2D Matrix Leetcode Solution.
Search a 2D Matrix Leetcode Solution in Python
class Solution:
def searchMatrix(self, matrix, target):
if not matrix or not matrix[0]: return False
m, n = len(matrix[0]), len(matrix)
beg, end = 0, m*n - 1
while beg < end:
mid = (beg + end)//2
if matrix[mid//m][mid%m] < target:
beg = mid + 1
else:
end = mid
return matrix[beg//m][beg%m] == targetSearch a 2D Matrix Leetcode Solution in CPP
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int n = matrix.size();
int m = matrix[0].size();
if(n == 0 || m == 0)
return false;
int start = 0, end = m*n - 1;
while(start <= end)
{
int mid = start + (end - start) / 2;
int ind = matrix[mid/m][mid%m];
if (target == ind)
return true;
else if(target < ind)
end = mid - 1;
else
start = mid + 1;
}
return false;
}
};Search a 2D Matrix Leetcode Solution in Java
class Solution {
public boolean bs(int[][] matrix,int row, int start, int end, int target){
while(start <= end){
int mid = start - ( start - end) / 2;
if(target == matrix[row][mid]) return true;
else if(target < matrix[row][mid]) end = mid-1;
else if(target > matrix[row][mid]) start = start + 1;
}
return false;
}
public boolean searchMatrix(int[][] matrix, int target) {
//finding potential row
int m = matrix[0].length-1;
int sRow = 0;
int eRow = matrix.length-1;
while(sRow <= eRow){
int mid = sRow - (sRow - eRow)/2;
if(target == matrix[mid][0] || target == matrix[mid][m]) return true;
else if(target > matrix[mid][0] && target < matrix[mid][m]){
return bs(matrix, mid, 0, m, target);
}
else if(target < matrix[mid][0]){
eRow = mid - 1;
}
else if(target > matrix[mid][m]){
sRow = mid + 1;
}
}
return false;
}
}Note: This problem Search a 2D Matrix is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.
